A) 32 cm
B) 40 cm
C) 28 cm
D) 36 cm
Correct Answer: A
Solution :
[a] Given that end correction |
\[e=1\,cm\]and shortest resonating length= |
\[10cm\] |
\[\frac{\lambda }{4}={{l}_{1}}+e\] |
\[\frac{\lambda }{4}=10+1=11cm\] |
\[\lambda =44cm\] |
Let the next resonating length be |
\[\frac{3\lambda }{4}={{l}_{2}}+e\] |
\[33={{l}_{2}}+1\] |
\[{{l}_{2}}=32\,cm\] |
Let the shortest resonating length be |
\[{{l}_{1}}\] |
And the next be\[{{l}_{2}}\] |
Then,\[{{l}_{1}}+e=\frac{\lambda }{4}\] ............ (1) |
\[{{l}_{2}}+e=\frac{3\lambda }{4}\] .............(2) |
Here, e = end correction |
On dividing (1) by (2) |
\[\frac{{{l}_{1}}+e}{{{l}_{2}}+e}=\frac{1}{3}\] |
\[\frac{10+1}{{{l}_{2}}+1}=\frac{1}{3}\] |
\[33={{l}_{2}}+1\] |
\[{{l}_{2}}=32cm\] |
the end correction given is 1 cm so the shortest length resonating will be one fourth of wavelength so the wavelength comes out to be 44cm the nest resonating length will be three fourth of wavelength so it will be 33cm and end correction given is 1cm so the next length resonating will be cm\[33-1=32\,cm\] so the correct option is A. |
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